JEE Main 2019 — Rotational Motion Question with Solution
From: JEE Main 2019 (Online) 12th January Evening Slot
Question
A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. when the particle reaches point b, its angular momentum about O will be :
(Take g = 10 m/s2)
(Take g = 10 m/s2)
This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: A
Correct answer
A6 kg-m2/s
Step-by-step explanation
Work Energy Theorem from A to B
Mgh = mv mv
2gh =
2 10 10 = v 52
vB = 15m/s
Angular momentum about 0
L0 = mvr
= 20 103 20
L0 = 6 kg.m2/s
Mgh = mv mv
2gh =
2 10 10 = v 52
vB = 15m/s
Angular momentum about 0
L0 = mvr
= 20 103 20
L0 = 6 kg.m2/s
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Rotational Motion chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2019, covering the Rotational Motion chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.